Prove that a straight line is the shortest distance between two points in $E_3$. Use the following scheme; let $\alpha: [a,b]\to E_3$ be an arbitrary curve segment from $p = \alpha(a) , q = \alpha(b)$.Let $u = (q — p)/||q — p || $.
(a) If $\sigma$ is a straight-line segment from $ p$ to $ q$ , say $$\sigma (t) = (1 - t)p + tq ,\quad 0\leq t\leq1$$ show that $L(\sigma ) = d(p,q)$.
What I have done $$ L(\sigma)=\int_{0}^{1}||\sigma'(t)||dt=\int_{0}^{1}(p^2+q^2)^{1/2}dt=\sqrt{(p^2+q^2)}(1), $$ $d(p,q)=\sqrt{(b-a)^2+(q-p)^2}$. Where am I doing wrong? It's a problem from O'Neill Elementary Differential Geometry.